Problem: Let $f(x, y, z) = xz + yz$ and $g(t) = (\cos(t), \sin(t), t)$. $h(t) = f(g(t))$ $h'(t) = $
Explanation: Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'(t) = \nabla f(g(t)) \cdot g'(t)$. $\begin{aligned} &g(t) = (\cos(t), \sin(t), t) \\ \\ &g'(t) = (-\sin(t), \cos(t), 1) \\ \\ &\nabla f = (z, z, x + y) \\ \\ &\nabla f(g(t)) = \left( t, t, \cos(t) + \sin(t) \right) \end{aligned}$ Substituting: $h'(t) = -t \sin(t) + t \cos(t) + \cos(t) + \sin(t)$ Answer $h'(t) = -t \sin(t) + t \cos(t) + \cos(t) + \sin(t)$